例 : 设 X , Y X,Y X,Y 相互独立 , X ~ P ( λ 1 ) X\sim P(\lambda_1) X~P(λ1?) , Y ~ P ( λ 2 ) Y\sim P(\lambda_2) Y~P(λ2?) , 求证 Z = X + Y Z=X+Y Z=X+Y 服从参数为 λ 1 + λ 2 \lambda_1 + \lambda_2 λ1?+λ2? 的泊松分布
证明 :
由题意 , X X X 的分布律为 P { X = i } = λ 1 i i ! e ? λ 1 , i = 0 , 1 , 2 , ? P\{X=i\}=\frac{\lambda_1^i}{i!}e^{-\lambda_1},i=0,1,2,\cdots P{X=i}=i!λ1i??e?λ1?,i=0,1,2,?
Y Y Y 的分布律为 P { Y = i } = λ 2 i i ! e ? λ 2 , i = 0 , 1 , 2 , ? P\{Y=i\}=\frac{\lambda_2^i}{i!}e^{-\lambda_2},i=0,1,2,\cdots P{Y=i}=i!λ2i??e?λ2?,i=0,1,2,?
Z Z Z 的可能取值为 0 , 1 , 2 , ? 0,1,2,\cdots 0,1,2,? , Z Z Z 的分布律为 P { Z = k } = P { X + Y = k } = ∑ i = 0 k P { X = i } P { Y = k ? i } = ∑ i = 0 k λ 1 i λ 2 k ? i i ! ( k ? i ) ! e ? λ 1 e ? λ 2 = e ? ( λ 1 + λ 2 ) k ! ∑ i = 0 k k ! λ 1 i λ 2 k ? i i ! ( k ? i ) ! = e ? ( λ 1 + λ 2 ) k ! ∑ i = 0 k C k i λ 1 i λ 2 k ? i = ( λ 1 + λ 2 ) k k ! e ? ( λ 1 + λ 2 ) P\{Z=k\}=P\{X+Y=k\}=\sum_{i=0}^{k}P\{X=i\}P\{Y=k-i\}=\sum_{i=0}^k\frac{\lambda_1^i \lambda_2^{k-i}}{i!(k-i)!}e^{-\lambda_1}e^{-\lambda_2}=\frac{e^{-(\lambda_1+\lambda_2)}}{k!}\sum_{i=0}^k\frac{k!\lambda_1^i \lambda_2^{k-i}}{i!(k-i)!}=\frac{e^{-(\lambda_1+\lambda_2)}}{k!}\sum_{i=0}^{k}C_k^i\lambda_1^i\lambda_2^{k-i}=\frac{(\lambda_1+\lambda_2)^k}{k!}e^{-(\lambda_1+\lambda_2)} P{Z=k}=P{X+Y=k}=∑i=0k?P{X=i}P{Y=k?i}=∑i=0k?i!(k?i)!λ1i?λ2k?i??e?λ1?e?λ2?=k!e?(λ1?+λ2?)?∑i=0k?i!(k?i)!k!λ1i?λ2k?i??=k!e?(λ1?+λ2?)?∑i=0k?Cki?λ1i?λ2k?i?=k!(λ1?+λ2?)k?e?(λ1?+λ2?)
k = 0 , 1 , 2 , ? k=0,1,2,\cdots k=0,1,2,?
类似地,可以证明, X ~ B ( n 1 , p ) , Y ~ B ( n 2 , p ) X\sim B(n_1,p),Y\sim B(n_2,p) X~B(n1?,p),Y~B(n2?,p) , 则 ? Z = X + Y ~ B ( n 1 + n 2 , p ) 则\,Z=X+Y \sim B(n_1+n_2,p) 则Z=X+Y~B(n1?+n2?,p)